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proof left as an exercise
07-01-2022, 07:51 PM (This post was last modified: 07-06-2022 01:33 PM by Albert Chan.)
Post: #10
RE: proof left as an exercise
Code:
sin((2*k-1)*x) / sin(x) = 1 + 2*[cos(2x) + cos(4x) + ... + cos((2*k-2)*x)]
sin(( 2*k )*x) / sin(x) = 2*[cos(x) + cos(3x) + cos(5x) + ... + cos((2*k-1)*x)]

Another way, using above sin(n*x)/sin(x) identities

1 + 4*cos(20°)
= 2*cos(20°) + 2*(cos(60°) + cos(20°))
= 2*cos(20°) + 4*cos(20°)*cos(40°)
= 2*cos(20°) * (2*cos(40°)+1)
= 2*cos(20°) * sin(60°)/sin(20°)
= 2*sin(60°) / tan(20°)

--> 2*cos(30°) / (1 + 4*sin(70°)) = tan(20°)

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Using complementary angle, we can get ratio of cos over cos (or, sin over cos)

XCAS> Q := makelist(k->f(k*x),1,4)
XCAS> normal(Q(f=sin, x=pi/2-y))      → [cos(y),sin(2*y),-cos(3*y),-sin(4*y)]
XCAS> normal(Q(f=cos, x=pi/2-y))      → [sin(y),-cos(2*y),-sin(3*y),cos(4*y)]

(x = pi/2-y) transform affected both side of identity; it is better to reverse sum order.
Code:
cos((2k-1)*y) / cos(y) = 2*[cos((2k-2)*y) - cos((2k-4)*y) + cos((2k-6)*y) - ... ] + (-1)^(k+1)
sin(( 2k )*y) / cos(y) = 2*[sin((2k-1)*y) - sin((2k-3)*y) + sin((2k-5)*y) - ...   + (-1)^(k+1)*sin(y)]

Example, redo the proof, using cos over cos identity:

cos(3x)/cos(x) = (2*cos(2x) - 1)

1 + 4*cos(20°)
= 2*cos(60°) + 4*cos(20°)
= 2*cos(20°)*(2*cos(40°)-1) + 4*cos(20°)
= 2*cos(20°) * (2*cos(40°)+1)
= 2*cos(20°) * sin(60°)/sin(20°)
= 2*sin(60°) / tan(20°)
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proof left as an exercise - Thomas Klemm - 06-06-2022, 11:41 PM
RE: proof left as an exercise - Albert Chan - 07-01-2022 07:51 PM



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